I have been trying to work out the answer to this question, but I don't have a markscheme so I'd like to confirm if I'm correct or not.

I drew out the number of orientations in which no two cars are parked next to each other and found that there are 6 combinations that fit the criteria.

I then found 6/(10C5), which gave me 6/252 or 1/42 as the probability. Is my working/answer correct?

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    $\begingroup$ You are correct in your answer to the question. $\endgroup$ – drhab May 12 '18 at 14:21
  • $\begingroup$ @NewGuy: They could be in $1,3,6,8,10$ as well, for example $\endgroup$ – Ross Millikan May 12 '18 at 14:22
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    $\begingroup$ Presumably, drivers arrive sequentially and each parks in a space chosen uniformly at random from the set of empty spaces? If you don't specify a probability distribution, the question of "probability" is not meaningful. You could instead ask a more purely combinatorial question, like "how many possible arrangements of cars are there, and how many arrangements are there in which no two cars are adjacent?". $\endgroup$ – wchargin May 12 '18 at 17:56
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    $\begingroup$ About 50%. Most drivers don't like to park adjacent to another car, so it's a tossup whether the second car will park in a space that permits an alternating pattern of empty and occupied spaces. If they do, adjacency aversion will take care of the rest. $\endgroup$ – Mark May 12 '18 at 20:34
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    $\begingroup$ In addition, the statement of the question doesn't define that these 10 empty spaces are all adjacent to each other. It could be a car-park that has a total of 1000 spaces, but only 10 are empty, none of which are adjacent. Then the answer is that there is a 100% chance of no two additional cars being parked next to each other. $\endgroup$ – Makyen May 12 '18 at 20:54

For a systematic approach:

Note that the possible patterns must have the form $$0^a\,X\,0^b\,X\,0^c\,X\,0^d\,X\,0^e\,X\,0^f$$

Where $0$ is an empty space, $X$ is a car, $a+b+c+d+e+f=5$ and $b,c,d,e>0$. Letting $b'=b-1,c'=c-1,d'=d-1,e'=e-1$ we see that $a+b'+c'+d'+e'+f=1$ so exactly one of these is non-zero (and that one is equal to $1$). That translates to $6$ options, confirming your solution.

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    $\begingroup$ Thank you for this; I got the combinations by drawing them out but I didn't know how to show it numerically. This really helps $\endgroup$ – user43712 May 12 '18 at 14:28
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    $\begingroup$ Worth noting: for more complicated set ups (with more spaces, for example), you can still do it this way though it will be harder to simply read off the answer. Stars and Bars helps a lot for those problems. $\endgroup$ – lulu May 12 '18 at 14:29
  • $\begingroup$ I think you should specify your pattern is a string, so people don't get confused with an algebraic expression. $\endgroup$ – qwr May 12 '18 at 21:08

Park $5$ cars on $6$ spots -- in $6=\binom{6}{5}$ ways. Add between each car one extra spot and get a proper parking on $6+4=10$ spots.

  • $\begingroup$ This is a really interesting way of approaching the question. Would this method work for 4 or 3 cars on a 10-space car park? $\endgroup$ – user43712 May 12 '18 at 18:23

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